Singularity Functions - Useful for beams with fixed ends? 2

[P1ENG]

I use singularity functions all the time for determining shear and moment diagrams for beams in MathCAD. My question is, can singularity functions be used when the beam has a fixed end(s)? I was assuming for a fixed end, that [θ]'=0 would be a boundary condition for solving for c1 and c2 in my [θ]' and [θ] equations, but I am not getting correct answers. I don't know if there is a mistake in my program somewhere, or if the mistake is my assumption of setting [θ]'=0 at the fixed ends.

Also, if anyone is familiar with both singularity functions and the moment distribution method, is there any advantage to the moment distribution method? I am not too familiar with that method, but I would think the only advantage is that the method can be applied to 2-D applications where singularity functions are limited to a continuous beam.

Juston Fluckey, E.I.
Engineering Consultant

 

[prex]

Not sure to understand what you are exactly doing: can you outline your procedure?
Taking the example given here Singularity function - Wikipedia, you would be required to add a constant of integration [tt]c[/tt] when calculating M : paraphrasing the example Because the moment is not zero at x=0, a constant of integration, c, is added (and of course end reactions need be left as unknowns and determined with [tt]c[/tt] from u'=0 and u=0 at the second end and from equilibrium).
Does this make sense?

prex

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[ishvaaag]

From what I see in this article

http://www.cgl.uwaterloo.ca/~tjlahey/sfunctions.pdf

the singularity functions are used mainly in the definition of the loading. Once that is made, they should be ready to stand any of the ordinary procedures of analysis (including any boundary conditions) analytical when feasible, and numerical where not. If I find any practical reference dealing with the amtter I will post it.

 

[LSPSCAT]

P1ENG,

Check out "Machine Design" by Robert L. Norton, pg 116-131 discusses the use of singularity functions for beam loading including examples of beams with fixed ends. Roark also tackles beam loadings with singularity functions. As ishvaag states, once the loading function is developed and you have integrated it is a matter of proper boundary conditions.

I am a mechanical who now mainly practices structural engineering so I was exposed to singularity funtions as an ME however I have found that it is never discussed in civil, including structural engineering grad classes that I have taken.

It is a great method for programming and graphically displaying beam loadings.

Good Luck! If you have a specific case post a sketch and we can work through.

 

[asixth]

Don't know MatchCad but what I programmed in Matlab was an if statememt when there are no degrees of freedom to skip the matrix imversion. You know the stiffness, load and displacememt so you can go strait to the post analysis.

 

[P1ENG]

I have some work to catch up on, but I will post a sample of my calculation soon. I fully understand how to use singularity functions as I too am an ME doing structural work. I don't want to look up conditions in Rourke's if I can do it all mathematically. The great thing about Singularity Functions is you can solve it even if you have a statically indeterminate situation. The thing is, once you add an unknown (moment due to fixed end condition), then you need to add an extra boundary condition in order to solve it. The definition of my boundary condition is what I am questioning.

Juston Fluckey, E.I.
Engineering Consultant

 

[ishvaaag]

Both slope and deflection are zero at fixity ends for fixed beams; this, of course, is not true (except for some cases) for intermediate support points where you have elastic fixity, as in supports in continuous beams or nodes at frames, where the joint itself gets some rotation from the undeformed shape to equilibrium. In other words, you only will state slope=0 where there is true fixity to rotation.

 

[ishvaaag]

Just to remark, the slope of the tangent at true fixity points is zero, so you can set slope=0 there.

 

[ishvaaag]

... for horizontal beams, of course (or relative to the axis of the beam, if straight).

 

[P1ENG]

ishvaaag, that was what I was assuming. I will have to go back and look at my program to see where I made the mistake.

Juston Fluckey, E.I.
Engineering Consultant

 

[rb1957]

are you looking at multi-span beams ? i'm not sure that you can define the boundary conditions of an individual span without solving the entire beam first (using MDM or other).

 

[P1ENG]

O.K., some time freed up. Take a look at my example (attached below) that will better explain what I am trying to do. My results are incorrect per the RISA results of the second page. At this point, I wouldn't mind if I made the dumbest, clearest mistake as long as someone can point it out. Thanks!

Juston Fluckey, E.I.
Engineering Consultant

 

[P1ENG]

prex! Going over the responses so far, I think I now understand what you were saying. I should introduce a constant, c, to my calculation of M(x). I was also thinking of doing this, but it leaves me with an extra unknown. I need an extra BC to be able to solve for it. Would that extra BC come from M(20')=0? I think so... I will redo my hand calculation and see what the outcome is!

Juston Fluckey, E.I.
Engineering Consultant

 

[P1ENG]

I corrected my example per my previous post, but I am still getting a different answer than the RISA results. See attached.

Juston Fluckey, E.I.
Engineering Consultant

 

[P1ENG]

IGNORE ALL MY PREVIOUS POSTS AND ATTACHMENTS, unless you want to read about my struggles of checking my work and thinking through the method.

The previous attachment integrated the c1 term incorrectly in the [δ](x) function. The attached hand calculation shows correct values. I don't know how RISA's solver handles the calculation of the reactions which are slightly different than mine, but I would think mine are 100% on.

Juston Fluckey, E.I.
Engineering Consultant

 

[ishvaaag]

I solve as you per RISA 3D (shear deformation influence disabled) and by the force method in Mathcad 2000 Professional using as sole unknowns M1, R1, R2 by your notation, and attach the results.

My E and I are different but for constant section irrelevant to moments and shears (not vertical deflections outside nodal restraints, or slopes anywhere). Both the Mathcad and RISA 3D solutions easily changeable to your E, I values.

In this case, I have not yet stated the problem solved in Mathcad by the force method with the singularity functions. All that is needed to do so is to include a previous definition of the singularity functions required in Mathcad, then modify the statement of the moment according to such function definition, to follow the same procedure to solution.

Since I don't know about the general use of the singularity functions for structural analysis, I can't see if there is the general convenience of retaining 7 unknowns where as here we can deal the matter (mostly) with 3. Since automatical solver, scarcely relevant at this scale.

A note for Mathcad users:
Note interestingly how the Mathcad vector was able to retain in Mathcad 2000 Professional solutions in different units (moments and vertical reactions). I have seen this kind of ability disabled in later releases.

For later relases it may turn necessary while solving to reduce to consistent (to be, implied) units and then recovery of the same, first by the corresponding division of the pertaining units of the component in the to be implied units and then multiplication of those corresponding after solution.

 

[P1ENG]

ishvaaag: Thanks for the calculation sample. I didn't think to use functions and the Find command in Mathcad. My approach involved using vectors. This will probably simplify my program very much and make it more versatile.

Also, my RISA results were for a W8x10. Making it a Rigid member yields the same results as ishvaaag. I wonder if the difference is due to shear deformation.

Next thought for everyone... what would happen if the right end was also fixed? You would then loose your known BC of M(20')=0.

Juston Fluckey, E.I.
Engineering Consultant

 

[prex]

You would then loose your known BC of M(20')=0.

But you replace it with y'(20')=0...

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

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: Air bearing pads

 

[ishvaaag]

If the right end is also fixed we lose the known moment quantity at the right support but gain there the slope=o known, so it can also be solved within the same setup, on only the same 3 unknowns than before.

I have not checked this time against RISA 3D.

 

[P1ENG]

Sleep deprivation... thanks prex. I am armed with a boundary condition arsenal to complete my program. I'm glad this all worked out.

Juston Fluckey, E.I.
Engineering Consultant